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(A) According to Einstein's photoelectric equation,the maximum kinetic energy $K$ is given by $K = E - \phi$,where $E$ is the photon energy and $\phi$

Vedclass · Accepted Answer

$2$

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(A) According to Einstein's photoelectric equation,the maximum kinetic energy $K$ is given by $K = E - \phi$,where $E$ is the photon energy and $\phi$ is the work function.For the first case: $K_{1} = E_{1} - \phi = 4 - \phi$.For the second case: $K_{2} = E_{2} - \phi = 2.5 - \phi$.Since $K = \frac{1}{2}mv_{max}^{2}$,the ratio of kinetic energies is $\frac{K_{1}}{K_{2}} = \left(\frac{v_{1}}{v_{2}}\right)^{2}$.Given $\frac{v_{1}}{v_{2}} = 2$,we have $\frac{K_{1}}{K_{2}} = 2^{2} = 4$,so $K_{1} = 4K_{2}$.Substituting the expressions for $K_{1}$ and $K_{2}$: $4 - \phi = 4(2.5 - \phi)$.$4 - \phi = 10 - 4\phi$.$3\phi = 6$.$\phi = 2 \ eV$.

The surface of a metal is illuminated alternately with photons of energies $E_{1} = 4 \ eV$ and $E_{2} = 2.5 \ eV$ respectively. The ratio of maximum speeds of the photoelectrons emitted in the two cases is $2$. The work function of the metal in $eV$ is:

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